Download Prof. Girardi Urysohn`s Lemma Urysohn`s Lemma is a crucial tool in

Prof. Girardi
Urysohn’s Lemma
Urysohn’s Lemma is a crucial tool in analysis. The aim of this handout is to present
Urysohn’s Lemma. To illustrate it’s power, we also present some of its corollaries.
Roughly speaking, Urysohn’s Lemma says that given two disjoint closed sets A and B
from a nice topological space, one can find a continuous function taking the value zero on
A and value one on B. To clarify nice, recall the separation properties, which give to what
extent points and subsets of a topological space can be separated.
Definition 1. Let X be a topological space.
(0) If for all x, y ∈ X with x 6= y there is an open set U such that x ∈ U and y ∈
/ U or
there is an open set V such that y ∈ V and x ∈
/ U , we call X a T0 space.
(1) If for all x, y ∈ X with x 6= y there is an open set U such that x ∈ U and y ∈
/ U and
there is an open set V such that y ∈ V and x ∈
/ U , we call X a T1 space.
(2) If for all x, y ∈ X with x 6= y there exist disjoint open sets U and V such that x ∈ U
and y ∈ V , the space X is called a T2 space.
(3) We say that X is a T3 space, provided it is T1 and satisfies the following condition.
For any closed set A and any point x ∈ X \ A there exist disjoint open sets U and
V such that x ∈ U and A ⊆ V .
(4) The space X is called a T4 space, provided it is a T1 space and fulfills the following.
For any disjoint closed sets A, B ⊆ X there exist disjoint open sets U and V such
that A ⊆ U and B ⊆ V .
A T2 space is also called a Hausdorff space, a T3 space is called a regular space, and a T4
space is called a normal space.
Clearly, a topological space X is a T1 space if and only if one point subsets of X are
closed. Thus the following implications hold.
(T4 ) ⇒ (T3 ) ⇒ (T2 ) ⇒ (T1 ) ⇒ (T0 ) .
It is not difficult to show that meterizable spaces and compact T2 spaces are normal spaces.
Theorem 2 (Urysohn’s Lemma). For a T1 topological space X, the following are equivalent.
(1) X is a normal space.
(2) For each pair of disjoint closed subsets A and B of X there exists f ∈ C(X, [0, 1])
with f (t) = 0 for each t ∈ A and f (t) = 1 for each t ∈ B.
The proof of Urysohn’s Lemma uses the following statement.
Lemma 3. Let X be a normal topological space. Let A and B be two disjoint closed subsets
of X. Define
∆ = k 2−n : 0 < k < 2n , n ≥ 1
(1)
to be the set of dyadic rational numbers in (0, 1). Then there is a family {Ur : r ∈ ∆} of
open sets in X such that
A ⊆ Ur ⊆ B c
for all r ∈ ∆ and
U r ⊆ Us
for all (r, s) ∈ ∆ × ∆ with r < s.
Proof of Lemma 3. Since X is normal, there exist two open sets V and W such that A ⊆ V
and B ⊆ W . We set U 1 = V . The closedness of W c and V ⊆ W c therefore imply
2
A ⊆ U 1 ⊆ U 1 ⊆ W c ⊆ Bc.
2
2
We now construct the sets Ur for r = k 2−n by induction on n. Set
∆N = k 2−n : 0 < k < 2n , 1 ≤ n ≤ N
and suppose that we have constructed open sets Ur with
A ⊆ Ur ⊆ U r ⊆ B c
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Prof. Girardi
Urysohn’s Lemma
for all r ∈ ∆N and
U r ⊆ Us
for all (r, s) ∈ ∆N ×∆N with r < s for an index N ∈ N. To get open sets Ur for all r ∈ ∆N +1
it remains to construct open sets Ur for r = (2j + 1) 2−(N +1) (0 ≤ j < 2N ). Using that
c
U 2j 2−(N +1) and U2(j+1)
are closed disjoint sets (where we set U 0 = A and U1c = B)
2−(N +1)
and that X is a normal space, we can choose an open set U(2j+1) 2−(N +1) with
A ⊆ U 2j 2−(N +1) ⊆ Ur = U(2j+1)2−(N +1) ⊆ U r ⊆ U2(j+1) 2−(N +1) ⊆ B c
for 0 ≤ j < 2N .
We inductively obtain a family of open sets {Ur : r ∈ ∆} which clearly has the desired
properties.
Proof of Theorem 2. Let X be a T1 topological space satifying property (2). Let A and B
be two disjoint closed subsets of X. Byproperty (2), there exists f ∈ C (X, [0, 1]) taking the
value 0 on A and 1 on B.
Then U := x ∈ X : f (x) < 13 is an open set containing A and
V := x ∈ X : f (x) > 32 is an open set containing B. Furthermore U and V are disjoint.
So X is normal.
To show that (1) implies (2), let X be a normal topological space. Let A and B be two
disjoint closed subsets of X. Let ∆ be as in (1) and let {Ur : r ∈ ∆} be the family of open
sets provided by Lemma 3. We set U1 = X and define
n
o
˜ : x ∈ Ur
f (x) = inf r ∈ ∆
˜ = ∆ ∪ {1}. We clearly have 0 ≤ f (x) ≤ 1 for all x ∈ X. Since
for x ∈ X, where ∆
c
A ⊆ Ur ⊆ B for all r ∈ ∆, we moreover deduce f (x) = 0 for all x ∈ A and f (x) = 1 for all
x ∈ B. It remains to show that f is continuous.
˜
To that purpose fix α ∈ R. We observe that f (x) < αSif and only if there is an r ∈ ∆
with r < α and x ∈ Ur , which is in turn equivalent to x ∈ r<α Ur . We conclude that
[
Ur
f −1 ((−∞, α)) =
r<α
is open.
Analogously, f (x) > α if and only if there is an r ∈ ∆ with r > α and x ∈
/ Ur . Here we
used that U σ ⊆ Uρ for all (σ, ρ) ∈ ∆ × ∆ with σ < ρ. This fact also allows us to deduce
that the last result is equivalent to the existence of a number s ∈ ∆ with s > α and x ∈
/ U s.
We conclude that
[ c
f −1 ((α, ∞)) =
U s,
s>α
−1
implying that f ((α, ∞)) is open. Since the open half-lines generate the topology on R,
we infer that f is continuous.
Urysohn’s Lemma is a key ingredient in the proofs of the following propositions.
Proposition 4 (Tietze’s Extension Theorem). Let X be a normal space. Let A be a closed
subset of X and f ∈ C(A, [a, b]). Then there exists a function F ∈ C(X, [a, b]) which
coincides with f on A.
Proof. Replacing f by (b − a)−1 (f − a), we may assume that [a, b] = [0, 1]. We inductively
construct a sequence (gn )n∈N of continuous functions on X such that
n
n
X
2n−1
2
0 ≤ gn ≤ n on X and 0 ≤ f −
gj ≤
on A
(2)
3
3
j=1
for all n ∈ N.
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Urysohn’s Lemma
Set B = f −1 ([0, 1/3]) and C = f −1 ([2/3, 1]). Since f is continuous on A, B and C are
closed in A. It follows that B and C are also closed in X as A is a closed subset of X.
Urysohn’s Lemma thus yields a continuous function g1 : X → [0, 1/3] with g1 = 0 on B and
g1 = 1/3 on C. In particular,
2
0 ≤ f − g1 ≤
3
on A. Now assume that we constructed continuous functions g1 , ..., gN on X such that (2)
P
is true for 1 ≤ n ≤ N . The function hN := f − N
j=1 gj is continuous on A so that
−1
−1
N
N +1
the sets BN := hN ([0, 2 /3
]) and CN := hN ([(2/3)N +1 , 1]) are closed in X by the
same reasoning as above. We again apply Urysohn’s Lemma to get a continuous function
gN +1 : X → [0, 2N /3N +1 ] with gN +1 = 0 on BN and gN +1 = 2N /3N +1 on CN . It follows that
N +1
2
0 ≤ hN − gN +1 ≤
on A.
3
P∞
Since kgn k∞ ≤ 2n−1 /3n for all n ∈ N, the series
n=1 gn converges uniformly to a
continuous function F on X. Moreover, we infer
n
2
on A
0≤f −F ≤
3
for all n ∈ N, implying f = F on A.
The proof of the next proposition shows that, for 1 ≤ p < ∞, if A is a compact subset
of RN and ε > 0, then there exists g ∈ Cc RN , [0, 1] with g agreeing with 1A on A and
kg − 1A kLp < ε.
Proposition 5. The space Cc RN , K of compactly supported continuous function from RN
to K is k·kLp -dense in Lp RN , K for 1 ≤ p < ∞.
Proof. Let 1 ≤ p < ∞. Since
)
( n
X
ai 1Ai : n ∈ N, ai ∈ K, Ai is a compact subset of RN
i=1
is k·kLp -dense in Lp RN , K , it suffices to show that
k·k
1A : RN → K | A is a compact subset of RN ⊂ Cc (RN , K) Lp .
(3)
Towards showing (3), fix a compact subset A of RN and ε > 0. Let Br (x) denote the open
ball in RN centered at x ∈ RN of radius r > 0. For each n ∈ N, let An := ∪a∈A B1/n (a).
Clearly each An is a bounded open subset containing A. Since A is closed, ∩∞
n=1 An = A.
Thus limn→∞ λ (An \ A) = 0. Choose n ∈ N such that λ (An \ A) < εp . Applying Urysohn’s
N
Lemma to the disjoint closed sets A and AC
n , find g ∈ C R , K taking values in [0, 1] and
C
taking the value 0 (resp. 1) on An (resp.
A). Then g is compactly supported since An is
bounded and kg − 1A kLp = g 1An \A Lp ≤ [λ (An \ A)]1/p < ε. Thus (3) holds.
References
[F] Gerald B. Folland, Real Analysis. Modern Techniques and Their Applications, Second edition. John
Wiley & Sons, Inc., New York, 1999.
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